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physics =(

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A person on a small, 7.00 meter high, hill aims a water-balloon slingshot straight at a friend hanging from a tree branch 4.00 meters above and 6.00 meters away. At the instant the water balloon is released at 7.00 m/s, the friend lets go and falls from the tree, hoping to avoid being hit by the water balloon. Calculate the following:
(a) the time for balloon to reach tree ______________s
(b) the height of balloon at that time ________________m
(c) the height of the falling friend at that time ___________________m


i don't even know where to begin. thanks!

  • physics =( - ,

    Draw a picture.
    It is not clear to me where the 7m hill comes in. If I interpreted it correctly, it is of no use except that it permits the water balloon and the person to fall freely below the ground!

    See:
    http://img689.imageshack.us/img689/5825/1260658746b.png

    Let
    D=horizontal distance between the two friends = 6m
    V=vertical distance between the two friends = 4m
    u= initial velocity of the balloon making an angle θ (theta) with the horizontal = 7 m/s
    ux=horizontal component of initial velocity = ucos(θ)
    uy=vertical component of initial velocity = usin(θ)

    H1(t)=height above initial level of balloon at time t
    = uy*t-(1/2)gt²
    H2(t)=height of friend measured from top of tree at time t
    = 0 -(1/2)gt²

    1. Time to reach tree
    T = D/ux
    2. H1(T)=uy*T-(1/2)gT²
    3. H2(T)=-(1/2)gT²

    Note that H1 is measured from ground, and H2 is measured from top of tree, at 4 m above ground.

    My calculations tell me that the balloon hits the poor friend.

  • physics =) - ,

    Forgot to mention:
    θ=(theta) is the angle aimed directly at the friend, so
    θ=arcTan(4m/6m)

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